In the book "Fundamentals of the Theory of Operator Algebras" by Richard V. Kadison and John R. Ringrose at page 111 (a screenshot is attached below) the authors claims:
Since the map $E \to I-E$ reverses the ordering of the projections we have $$\vee(I-E_a)=I-\wedge E_a$$
I have managed to prove $\vee(I-E_a) \le I-\wedge E_a$:
By definition, $\forall a, E_a \ge \wedge E_a \implies \forall a, I-E_a \le I - \wedge E_a$ (since $E \to I-E$ is reversing)$\implies \vee(I-E_a) \le I-\wedge E_a$
But I am not able to see how does the fact that "the map $E \to I-E$ reverses the ordering" implies the other inequality i.e. $\vee(I-E_a) \ge I-\wedge E_a$.
Any hint or suggestion is highly appreciated. Thanks in advance.
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You have $I-E_b\leq \lor(I-E_a)$. Then
$$ I-\lor(I- E_a)\leq E_b. $$ As this holds for all $b$, $$ I-\lor(I- E_a)\leq \land E_a, $$ which is $$ I-\land E_a\leq\lor(I-E_a). $$